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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution</dfn></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_15.html">
\begin{equation*}
\begin{aligned}
&amp;\frac{\textrm{d} y}{\textrm{d} x}=\frac{2 x}{(1+x^2) y}~\rightarrow~y \textrm{d} y=\frac{2 x}{1+x^2} \textrm{d} x,\\
&amp;~\rightarrow~\frac{1}{2} y^2=\ln (1+x^2)+C~\rightarrow~y^2=2 \ln (1+x^2)+C,\\
&amp;~\rightarrow~y=\pm \sqrt{2 \ln (1+x^2)+C}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Note, this ODE has two general solution, i.e., the solution is not unique. Further, to determine the integration constant, we use <span class="process-math">\(y(0)=-2\)</span> and have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_15.html">
\begin{equation*}
-2=\pm \sqrt{2 \ln (1+0)+C}.
\end{equation*}
</div>
<p class="continuation">We must take <span class="process-math">\(``-"\)</span> on the right hand side for the present problem and this gives <span class="process-math">\(C=4\text{.}\)</span> Thus, the solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_15.html">
\begin{equation}
y=-\sqrt{2 \ln(1+x^2)+4}.\tag{2.3.3}
\end{equation}
</div>
<p class="continuation">Note, for the ODE, we require <span class="process-math">\(y \neq 0\text{.}\)</span> For (<a href="" class="xref" data-knowl="./knowl/eq2_15.html" title="Equation 2.3.3">(2.3.3)</a>), <span class="process-math">\(y\)</span> is less than zero and can never be zero for all <span class="process-math">\(x\text{.}\)</span> Thus, it is valid in <span class="process-math">\(-\infty \leq x \leq \infty\text{.}\)</span></p>
<span class="incontext"><a href="sec2_3.html#p-29" class="internal">in-context</a></span>
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